 Hello everyone. Last lectures, we discussed about the radioactivity and the decay, radioactivity decay and also the nuclear structure and stability and the properties of nuclei. Now, whenever we have a system and we have it has got certain characteristics, it is required to make a model that will explain the properties of the system. So, in the case of nucleus, there are many properties that nuclei possess and we need to have a model which will explain almost all properties of this nuclei. So, today we will start discussing the different models that are proposed and we discussed them how these models work and how they explain the properties of nuclei. So, I will just give a concise slide depicting different models that have been proposed to explain the nuclear structure, nuclear properties and so on. So, the first one is the liquid drop model which in fact explains the gross property of the nuclei and then there is another model called shell model which explains the fluctuations in the gross properties of nuclei. So, as we discussed, you will find that liquid drop model gives you the gross properties, the shell model explains the fluctuations in those properties and there is a synthesis of these two types of models to give you empirical shell models wherein you use the experimental data to generate the properties of nuclei. So, there are the masses like masses of nuclei with empirical models and there are many body calculations from the first principles you can try to calculate the properties of nuclei. So, this is regard to the structure and properties. In nuclear reactions, when we are studying the nuclear reactions, there is another model called as the optical model. There are also some of other models called collective models which explain the collective properties of nuclei, ribotation, vibration, etc. We will be concentrating only on the liquid drop model and the shell model in this particular course. So, let me first start with the liquid drop model. The liquid drop model was proposed by Niels Bohr in 1930s and in fact, he was trying to explain the nuclear fission process, how to explain the splitting of a nucleus, heavy nucleus particularly into two fragments of nearly equal mass and he thought of a concept like it will have a drop of liquid and if you try to reform it, how it can split into two droplets, that is how liquid drop gave the concept of the liquid drop. So, we have discussed the different properties of nuclear matter and we may use of those properties to build up this liquid drop. So, in the nuclear structure we found that nuclear matter has got very high density. If you recall the previous lecture, we discussed that the density of nuclear matter is 14 grams per cc or 10.38 nucleons per centimeter cube. So, that is like nuclear matter is highly incompressible. So, it is envisaged in the liquid drop model that the nucleus is like a drop of incompressible matter. We also saw that the density the central density of all nuclei is same and that is of the order of 10 power 38 nucleons per cc. So, all nuclei have the same density then nuclear force, nuclear force is short range and attractive. What are the evidences for that? The short range property of nuclear force, the evidence comes from the saturation property of binding energy. So, binding energy per nucleon is constant and so, if it was not constant, if each nucleon is interacting with all nucleons in the nucleus then the binding energy per nucleons would be proportional to the mass number. But actually the binding energy per nucleons constant means the binding the nucleons are interacting only with those nucleons which are immediate vicinity. It is attractive force, it is a imperative that you want to break the nucleus into constituent nucleus where to supply energy. So, the nucleons are tightly held together and which works at a very short range. Nuclear force is saturated we already discussed and so, it is like you know in a the Van der Waals forces between the molecules is just like a very short range force. So, that is how we envisage that the nuclear force is like a Van der Waals force between molecules in a drop of liquid. And then the in that analogy the binding energy is analogous to the heat of aberration of liquid drops. If you want to remove a nucleus nucleon from the nucleus which is like evaporating a drop you know a part of the liquid. So, there are there are the analogies and we will build the nuclear liquid drop model based on these properties and these assumptions. So, the important part of liquid drop model is lies in its success in predicting the masses of the nuclei. So, that is a very important success of liquid drop model and our focus will be more on the calculating the binding energy of the nucleus and thereby the nuclear mass. So, this semi empirical mass formula by semi empirical essentially we mean that it is based on some phenomenology analogous to liquid drop model but the constants of that liquid drop model mass formula are obtained empirically using the experimental masses of nuclei and just proposed by von Weijker in 1935. So, in the in the liquid drop model the most important thing is to predict the mass of a nucleus or essentially which requires the predicting the binding energy of it. So, we are focusing more on binding energy of the nuclei. So, the binding energy is essentially given in terms of the mass of the nucleus equal to mass of the z protons in the nucleus and the mass of the a minus z neutrons in the nucleus. So, when we combine z protons and a minus z neutrons to form a nucleus of mass m, then energy equivalent to the binding energy is released. In other words, if you want to break this nucleus into these two constituents, concentrated nucleons, then with the energy we have to supply energy equivalent to the binding energy of nuclei. So, this binding energy is the term which we will focus in this permanent lecture and the binding energy B is constituted of several terms which we call as volume energy BV, the surface energy BS, the Coulomb energy BC, the asymmetry energy BA and the pairing energy BP. So, these are the all the volume the surface are the properties of the liquid drop. So, we derive the expressions of the volume energy and surface energy from the liquid drop properties and then to that there are some additional terms which depending upon the they like the nucleus instead of the liquid drop the nucleus contains charge and the charges are repelled like the other repelled. So, that is how the Coulomb term came into picture and the neutrons and protons are not same particularly for the heavy nuclei and so we have the asymmetry term and the pairing energy term that means the nucleons like to pair up inside the nucleus. So, there is a pairing energy term. So, let us see the how to construct the semi empirical mass problem. So, we know that the mass of the nucleus is proportional to the mass number because the mass numbers mass number is nothing but proton number per neutron number. So, mass is very close to the mass number there will be slight difference between mass actual mass and the mass number and the banding energy is the energy when the nucleons combine to form the nucleus then the energy is released. So, the volume and mass are related to the density and the density of nuclear matter is constant. So, if you take a infinite nuclear matter we know that the nucleons interact with the other nucleons then for the infinite matter if you take the n nucleons then the binding energy is proportional to n because we know that the average binding energy is constant b proportional to a or b by a is constant. So, that is what we have tried to generate here that the b the binding energy of nucleus is proportional to mass number and this proportionality constant nothing but the this term what is that the volume term. So, the volume is 4 by 3 pi r cube which is nothing but proportional to mass number because density is constant. So, binding energy you can write as proportional to the volume term 4 by 3 pi r square into a. So, this should have been r cube so that is equal to a and this so 4 by 3 pi it is known quantity r 0 is known radius constant. So, this 4 by 3 pi r 0 is termed as the a v the volume energy constant. So, this volume energy essentially constitutes the binding energy of a infinite nuclear matter if there were no boundaries of the nuclear matter then the binding energy per nucleon is constant and we can say that like we have this binding energy per nucleon binding energy per nucleon it should be constant. So, that is what is the per nucleon but if you take the binding energy then it should have been like this because this the slope is nothing but a v. So, binding energy per nucleon is constant because the binding energy is equal to a v into a. Now, this is a very hypothetical situation the nuclear nucleus is not infinitely large in size it has got certain size because there are finite number of nucleons that is where if it is not infinite then it has got a surface. So, the surface energy term in fact is like a correction to the volume energy term. So, those nucleons which are at the surface of the nucleus so they do not have the neighbors on the outside. So, they are pulled up by the nucleons inside their nucleus. So, this is like you know the surface tension force that is how a drop of liquid tries to retain a surface as a spherical shape because this surface tension is minimum for a spherical shape. So, the surface area tries to minimize because the surface tension is proportional to the surface area. So, if you see the surface energy is nothing but the surface area proportional surface area of the nucleus and surface area is 4 pi r square you can again write r equal to r0 a raise to one-third. So, it becomes 4 pi r0 square a to the power two-third. So, now you see here the nucleons at the surface their contribution has already been taken into consideration in the volume energy term. The volume energy term takes care of all nucleons in the nucleus assuming that there is no surface. So, out of those a nucleons there are a few nucleons which are at the surface. So, there is a negative correction to the volume energy because of the nucleons at the surface. That is why this surface energy term is a negative contribution to binding energy minus of a s the surface energy constant into a to the power two-third because the surface area is proportional to r square and hence a raise to two-third. Now, you see a to the power two-third if you see this was the AV term this was the AV term then the surface energy term will become like the surface energy term will be at low mass all the nucleons are will be at the surface. So, as you go to higher and higher mass there will be less and less nucleons at the surface. So, the surface energy term is dominating at lower masses and becomes less and less as you go to higher masses. Third term is the the coulomb energy term and this coulomb energy term is coming into this binding energy because the proton-proton are repelling each other and so they will try to de-stabilize their nucleus. So, they would like to be away and so the coulomb energy term we have to there is again the subtraction to the volume energy term. Now, how to calculate the coulomb energy? The coulomb energy of a sphere suppose there are z protons each proton will interact with the remaining z minus 1 protons. So, the coulombic energy of a sphere containing z charges is z into z minus 1 e square upon r where r is the radius of the sphere. This is proportional to AVc, Vc is the coulomb energy term. So, you can now substitute for r r 0 a raise to one-third. So, it becomes and particularly when z is very large compared to 1 then z z minus 1 is z square. So, it becomes z square e square upon r 0 e to the power one-third. So, the coulomb energy term becomes minus ac where ac is nothing but e square upon r 0 and it becomes z square upon e to the power one-third. So, this coulomb energy term again it is a negative contribution to binding energy because the this is the ripple skip term. The volume energy term is attractive term but the coulomb energy term is a ripple symptom. So, it is trying to decrease the binding energy of the nucleus. Then the so the coulomb the the coulombic term you see if you see z square by a-third so for the lower mass region the z is low the contribution of coulombic term will be much less. So, if you add to this so actually by surface energy energy volume binding energy has decreased now the coulomb energy will try to become see here will be dominating for the heavy nuclei. So, coulomb energy will be more for the higher z nuclei and for low low mass number it is less. The origin of asymmetry energy term lies in the excess neutrons that we need to add to stabilize a nucleus particularly for those nuclei which have atomic number more than 20. So, you know for calcium for up to calcium for t z equal to 20 n equal to z means the number of neutrons if it is equal to number of protons then it is succeeded to stabilize the nuclei. But for nuclei having atomic number more than 20 then we need to have more neutrons to stabilize nucleus because we need to compensate for the coulombic repulsion between the protons. Now the origin of this term lies in the if you see the potential energy term the potential for the protons and neutrons. So, the protons suppose they are lying in a square level potential then proton will have a coulomb barrier and so these are the proton levels. So, and this is the Fermi level this is the binding energy of. So, this is the minus V0 for protons. Now if you have more number of neutrons then the Fermi level for proton and neutron has to be same otherwise it will become it will start going beta minus decay. So, to accommodate the more number of neutrons than protons and have the same Fermi level that neutron potential well will start from here. So, it is minus V0 for. So, there are some neutrons which are excess over protons and they do not have the benefit of exchanging with the protons and so this excess neutrons this excess neutrons in fact have higher kinetic energy compared to the protons. So, there will be some fraction of neutrons which will be having higher kinetic energy than protons and hence the lesser binding energy than the other other nucleus. So, that is the deficit in binding energy due to asymmetry of neutrons over the protons. Now how to calculate this term the asymmetry energy is proportional to the excess neutrons n minus z into the volume the fraction fraction of the volume occupied by the excess neutrons that is n minus z n minus z upon a that the fraction of the volume occupied by this n minus z neutron. So, excess neutron into the fractional volume is equal to the asymmetry energy and if you remove the proportionality constant it becomes a a n minus z square upon a to the power it is not a to the power 1 such it is a n minus z upon square upon a you can also write this as a minus 2 z square upon a in a. So, this is the binding energy term corresponding to asymmetry energy and its origin lies into the excess neutrons that are in the nucleus to stabilize that. So, their contribution has already been taken care in the average the volume energy term, but because they are excess over neutrons and they that fraction is actually not having the benefit of binding having exchange with the other protons. So, that is the origin of acidity. The last one is the pairing energy as you know the nucleons tend to pair up and so as a matter of convention for odd a nuclei the pairing energy has been taken as 0 if it is a if it is a odd or ordered nucleus then the they are binding energies low. So, the pairing energy minus delta minus delta and if it is a even even nucleus then the binding energy is high. So, for them the binding energy is plus 2. So, that is the convention we will use for pairing energy term. So, let us recapitulate whatever we have discussed in this slide. So, the liquid drop model using the semi empirical mass formula gives a the calculation of the binding energy in terms of different terms like volume term, surface term, pull on term, asymmetry term and pairing energy. So, you can write if you recall the previous expressions that we have been deriving is the binding energy of a nucleus is equal to A v into A, A is the mass number minus A s upon A raise to two-third the surface energy term minus A c z square by A raise to one-third the preliminary term minus A a and minus z square upon A, A a is the mass in asymmetry term and plus minus delta could be plus minus or 0 depending upon the mass number. Now, this the constants what are there are five constants A v, A a, A c, A a, A v, A s, A c, A a and delta these five constants have been determined empirically from experimentally known masses. Masses have been obtained using mass spectrometers and different other formulations. So, if you have the experimental masses, put them into this equation and do the fitting to get these five constants that is how we get we call it the empirical, empirically we have obtained these masses hence the name semi empirical markdown. A form life based on some phenomenology, but the constants are obtained by empirically. So, these are the terms A v is not in the small a, the volume energy constant is 14.1 ammv, surface energy constant 13 ammv, volume energy constant 0.595 ammv. Like you see here the it is all binding energy that the other there is the other things are constant. So, it means other things are not having any dimensions. So, the dimensions of energy coming from these constants only as symmetry term constant 19 ammv and delta the pairing energy 1.2 to 2 ammv. So, now, these five parameters in fact, they have been undergoing lot of refinement over the last many many years and Meyer's and Swatecky in 1966 gave a what is a very enhanced having enhanced prediction of the nuclear masses. So, Meyer's and Swatecky's are now they are known to develop the liquid drop model to a greater extent because the constant that had been obtained now they are they owe them to Meyer's and Swatecky. So, these constants in in terms of the five constants can calculate the masses of nuclei and they give a remarkably good representation of the average masses of nuclei over the entire mass region. What I try to explain here in the graph that the this noise what we plot the average binding energy the volume energy term av is constant about 14.1 ammv. So, you can put here 14.1 and the surface energy term is dominating at lower mass at higher mass becomes less the Coulombian term is very small at low mass but higher mass it becomes dominating. So, it is decreasing the binding energy and acidity again becomes important above the mass number 40. So, as for higher nuclei it becomes important. So, you can see here by the liquid drop mass we are able to generate the nature of the binding energy per nucleon. We recall the binding energy curve in the previous lectures this is the it is rising in the beginning and falling in the end becoming maximum at mass number around 60. So, liquid drop model successfully explains the trend of the average binding energy and these constants are used to find out the masses of the nucleus. So, before I go into the complete applications of liquid drop in different areas just to one example of the success of the liquid drop model is in predicting the atomic masses and their binding energies. So, you see here the binding energy now you can substitute in the mass the mass of the nucleus. So, z protons plus n a minus z newtons minus binding mass of the nucleus is equal to binding energy or you want to calculate the mass of the nucleus you can put into this formula mass of z protons minus plus mass of a minus z newtons minus binding energy. So, you calculate the binding energy by this formula and substitute in this equation we know the mass of proton and neutron can calculate nuclear mass. So, you can now try to explain to put this try to make it do some calculations here. So, binding energy equal to you can write now here a into m n this term plus z into m h minus m n you can take from here minus you just do some jugglery minus a plus a. So, you add a 1 a and minus 1 a simplify things. So, you can now take binding energy per nucleon divided by a experimental m n minus 1 m n minus m h z by a and minus m minus a upon a. So, this is you can write m minus a upon a this a has been taken here. So, now you can calculate the binding energy by this formula a v a a raise to 2 a s raise to 2 third minus a c z square by a raise to 1 third minus a a a minus 2 z. So, a minus z and n minus z can also be product a minus 2 z equally n minus z you can also call a minus z minus z, but that is nothing but a minus 2 z square upon a. So, this is the calculated binding energy using these constants and so, you can write now this constant and now what I am showing in the table here is the calculated and experimental masses. So, you can calculate experimental binding energy from here and experimental can calculate from here see the table here now for nuclei of different mass numbers. So, option 17 see the volume energy term is constant 14.1 a v surface energy term is changing now it is decreasing with the mass number volume energy term is increasing with the mass number estimating energy then the binding energy per nucleon calculated and experimented. So, what you can see here it is very accurate quite accurately predicting the masses the or the binding energy. So, once you know binding energy you can calculate the masses. So, they calculated and the experimental binding energies are very close to each other. So, this explains the success of the liquid drop form model in explaining the masses of the nucleon. So, this in fact is the success of the liquid drop model and I will stop here and then subsequently I will discuss the applications of liquid drop model other application liquid drop model. Thank you very much.